I'm trying to do the following exercise from Rudin's Real and Complex Analysis:
Suppose $f\in C(T)$ and $f\in \text{Lip }\alpha$ for some $\alpha>0$. Prove that the Fourier series of $f$ converges to $f(x)$, by completing the following outline: It is enough to consider the case $x=0$, $f(0)=0$. The difference between the partial sums $s_n(f;0)$ and the integrals
$$\frac{1}{\pi}\int_{-\pi}^{\pi}f(t)\frac{\sin nt}{t}\:dt$$
tends to $0$ as $n\to\infty$. The function $f(t)/t$ is in $L^1(T)$. Apply the Riemann-Lebesgue lemma. More careful reasoning shows that the convergence is actually uniform on $T$.
I already know how to prove the result as in Lipschitz Continuity and Hölder Continuity helps Fourier series to converge.
However, I don't understand Rudin's outline.
- When he says that it is enough to consider the case $x=0$, $f(0)=0$, is it simply because it changes almost nothing to the proof?
- Since we can prove that $s_n(f;0)$ tends to $0$ by proving that $f(t)\cot(t/2)$ is in $L^1$, it seems to me that Rudin's outline is only better if it is easier to prove that $f(t)/t\in L^1$ and that the difference between $s_n(f;0)$ and the integrals tends to $0$. But I don't see any easy way to do the latter.
- Why is the convergence uniform and why does it matter?
I would appreciate if anyone could answer those 3 questions.
Best Answer
The aim is to show that $s_n(f,a) \rightarrow f(a) $ for all $a$, however it is easy to show that if you let $g(x)=g(x+a)-g(a)$, $$s_n(f,a)-s_n(g,0) = \sum_{-n}^n \frac1{2\pi}\int_{-\pi}^\pi (f(t)e^{ik(a-t) } - g(t)e^{-ikt})dt = f(a)$$ so if you showed that $s_n(g,0)\rightarrow 0$ you may deduce that $s_n(f,a)\rightarrow f(a)$.
I think I agree with what you say. As $$s_n(f,0)=\frac1{2\pi}\int_{-\pi}^\pi f(t)\left(\sum_{-n}^n e^{-ikt}\right)dt = \frac1{2\pi}\int_{-\pi}^\pi f(t)\left(\sin(nt)\cot(t/2)+\cos(nt)\right)dt.$$ There is no additional complexity in showing that $t\mapsto f(t)\cot(t/2)$ is $L^1$ compared to $t\mapsto f(t)/t$, and then the result comes directly from Riemann-Lebesgue lemma. My only guess is that Rudin might be implicitly referring to the approximation of the Dirichlet kernel by the sinus cardinal function: $$D_n(x)=\frac{2\sin (nx)}x+\left(\sin (nx) \left(\operatorname{cotan} \frac x 2-\frac2x\right)+\cos(nx)\right)$$ where $D_n(t)=\sum_{-n}^n e^{-ikt}$.
Uniform convergence is much stronger than pointwise convergence, it definitely makes a difference!